Question

Given `1.50*mol*L^-1` `HCl`, what volume will be required for equivalence with a `3.21*g` mass of `Mg(OH)_2`?

Answer 1

Your question is unsound....

Explanation:

Magnesium hydroxide will probably not be soluble in water to the extent of `1.0*mol*L^-1`...let us modify the question.....and specify a mass of .........

`1.0*mol*L^-1xx55.0xx10^-3*Lxx58.32*g*mol^-1=3.21*g`

..i.e. a molar quantity of `0.055*mol` with respect to `Mg(OH)_2`.

This approach is not too abstract, in that you can commonly buy so-called `"milk of magnesia"`, a suspension of magnesium hydroxide in water that you consume if you got a dodgy tummy.

And we write (as a necessity) the stoichiometric equation....

`Mg(OH)_2(s) + 2HBr(aq) rarr MgBr_2(aq) + 2H_2O(l)`

And this shows us that TWO EQUIV of acid are required for neutralization.

And so we need .............

`(0.055*molxx2)/(1.50*mol*L^-1)xx10^3*mL*L^-1=73.3*mL`

...with respect to the acid supplied.....